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 August 15th, 2018, 06:32 PM #1 Senior Member   Joined: Jan 2012 Posts: 137 Thanks: 2 Evaluating a limit Hi, Please evaluate the following limit: $\displaystyle \lim_{x->0}\left ( \tfrac{1}{x^{2}}-\cot^{2}x \right )$ Thanks. Last edited by skipjack; August 15th, 2018 at 07:41 PM. August 15th, 2018, 07:03 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 621 Thanks: 394 Math Focus: Dynamical systems, analytic function theory, numerics The limit is 2/3. August 15th, 2018, 07:27 PM   #3
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Quote:
 Originally Posted by SDK The limit is 2/3.

Last edited by skipjack; August 15th, 2018 at 08:29 PM. August 15th, 2018, 07:44 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,663 Thanks: 2643 Math Focus: Mainly analysis and algebra Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$ setting $2t=x$ we have that \begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\ 4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ 3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\ &=2 \\ L&=\frac23 \end{align} Thanks to this for inspiration. Thanks from happy21 and topsquark August 15th, 2018, 08:21 PM   #5
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 Originally Posted by v8archie Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$ setting $2t=x$ we have that \begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\ &= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\ 4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ 3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\ &=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\ &= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\ &=2 \\ L&=\frac23 \end{align} Thanks to this for inspiration.
Brilliant! This method is very nice though uncommon, I think!

Thx. August 15th, 2018, 09:06 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,747 Thanks: 2133 \displaystyle \begin{align*}\cot^2\!x &= 1/(\sin^2\!x) - 1 \\ &= 1/(x^2(1 - x^2/3 \,+\; ...)) - 1 \\ &= (1/x^2)(1 + x^2/3 \,+\; ...) - 1 \\ &= 1/x^2 - 2/3 \,+\; ...\end{align*} so the required limit is 2/3. Thanks from happy21 August 16th, 2018, 05:06 AM   #7
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Quote:
 Originally Posted by happy21 Can you please explain?
If you show me some work I'll show you some work. August 16th, 2018, 09:59 AM   #8
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 Originally Posted by SDK If you show me some work I'll show you some work.
Yeah I did from the methods depicted above. Tags evaluating, limit Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mike1127 Calculus 6 March 22nd, 2016 06:14 PM rasa97 Calculus 5 August 28th, 2015 03:42 PM szz Calculus 1 December 7th, 2014 04:17 PM Aurica Calculus 2 June 21st, 2009 06:34 PM lefucer Calculus 2 June 28th, 2008 05:11 PM

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